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Factoring Trinomials with Leading Coefficient Other Than 1

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken great care in developing the student's understanding of the factorization process. The technique is consistently illustrated by displaying an empty set of parentheses and describing the thought process used to discover the terms that are to be placed inside the parentheses. The factoring scheme for special products is presented with both verbal and symbolic descriptions, since not all students can interpret symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with leading coefficients different from 1. Objectives of this module: be able to factor trinomials with leading coefficient other than 1.

Overview

  • The Method of Factorization

The Method of Factorization

In the last section we saw that we could easily factor trinomials of the form x2+b x+c x 2 b x c by finding the factors of the constant c c that add to the coefficient of the linear term b b, as shown in the following example:

Factor x2 - 4 x - 21 x 2 - 4 x - 21 .
The third term of the trinomial is −21 −21 . We seek two numbers whose

(a) product is −21 −21 and
(b) sum is −4 −4 .

The required numbers are −7 −7 and +3 +3 .

x 2 - 4 x - 21=( x - 7 ) ( x+3 ) x 2 - 4 x - 21 ( x - 7 ) ( x 3 )

The problem of factoring the polynomial a x 2 +b x+c a x 2 b x c , a≠1 a≠1, becomes more involved. We will study two methods of factoring such polynomials. Each method produces the same result, and you should select the method you are most comfortable with. The first method is called the trial and error method and requires some educated guesses. We will examine two examples (Sample Sets A and B). Then, we will study a second method of factoring. The second method is called the collect and discard method, and it requires less guessing than the trial and error method. Sample Set C illustrates the use of the collect and discard method.

The Trial and Error Method of Factoring ax2+bx+cax2+bx+c

Trial and Error Method

Consider the product

Steps showing the product of two binomials 'four x plus three,' and 'five x plus two.' See the longdesc for a full description.

Examining the trinomial 20 x 2 +23 x+6 20 x 2 23 x 6 , we can immediately see some factors of the first and last terms.

Table 1
20 x 2 20 x 2 6
20x,  x 20x,  x 6,  1 6,  1
10x,  2x 10x,  2x 3,  2 3,  2
  5x,  4x   5x,  4x  


Our goal is to choose the proper combination of factors of the first and last terms that yield the middle term 23 x 23x.
Notice that the middle term comes from the sum of the outer and inner products in the multiplication of the two binomials.
 The product of two binomials four x plus three, and five x plus two. The outer product of binomials is eight x, and the inner product is fifteen x.
This fact provides us a way to find the proper combination.

Look for the combination that when multiplied and then added yields the middle term.

The proper combination we're looking for is

The product of the first and the last term is twenty x squared. One of the combinations of the factors of the first and last term yields two new factors of the product such that their sum is the middle term: twenty three x.

Sample Set A

Example 1

Factor 6 x 2 +x−12 6 x 2 +x−12 .

Factor the first and last terms.

The factors of the first term 'six x squared' and the last term 'negative twelve' are shown. The product of the first and the last term is negative seventy-two x squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: x.

Thus, 3x 3x and 3 are to be multiplied, 2x 2x and −4 −4 are to be multiplied.

6 x 2 +x−12 = ( )( ) Put the factors of the leading term in immediately.   = ( 3x )( 2x ) Since 3x and 3 are to be multiplied, they must be located in different binomials. = ( 3x )( 2x + 3 ) Place the −4 in the remaining set of parentheses. = ( 3x − 4 )( 2x + 3 ) 6 x 2 +x−12 = ( 3x − 4 )( 2x + 3 ) 6 x 2 +x−12 = ( )( ) Put the factors of the leading term in immediately.   = ( 3x )( 2x ) Since 3x and 3 are to be multiplied, they must be located in different binomials. = ( 3x )( 2x + 3 ) Place the −4 in the remaining set of parentheses. = ( 3x − 4 )( 2x + 3 ) 6 x 2 +x−12 = ( 3x − 4 )( 2x + 3 )

Check:   (3x−4)(2x+3) = 6 x 2 +9x−8x−12 = 6 x 2 +x−12 Check:   (3x−4)(2x+3) = 6 x 2 +9x−8x−12 = 6 x 2 +x−12

Example 2

Factor 8 x 2 −30x−27 8 x 2 −30x−27 .

Find the factors of the first and last terms.

The factors of the first term 'eight x squared' and the last term 'negative twenty seven' are shown. The product of the first and the last term is negative two hundred sixteen x squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: negative thirty x.

Thus, the 4x 4x and −9 −9 are to be multiplied, and 2x 2x and 3 are to be multiplied.

8 x 2 −30x−27 = ( 4x )( 2x ) = ( 4x )( 2x − 9 ) = ( 4x + 3 )( 2x − 9 ) 8 x 2 −30x−27 = ( 4x )( 2x ) = ( 4x )( 2x − 9 ) = ( 4x + 3 )( 2x − 9 )

check:   (4x+3)(2x-9) = 8 x 2 −36x+6x−27 = 8 x 2 −30x−27 check:   (4x+3)(2x-9) = 8 x 2 −36x+6x−27 = 8 x 2 −30x−27

Example 3

Factor 15 x 2 +44x+32 15 x 2 +44x+32 .

Before we start finding the factors of the first and last terms, notice that the constant term is +32 +32 . Since the product is positive, the two factors we are looking for must have the same sign. They must both be positive or both be negative. Now the middle term, +44x +44x , is preceded by a positive sign. We know that the middle term comes from the sum of the outer and inner products. If these two numbers are to sum to a positive number, they must both be positive themselves. If they were negative, their sum would be negative. Thus, we can conclude that the two factors of +32 +32 that we are looking for are both positive numbers. This eliminates several factors of 32 and lessens our amount of work.
 Factor the first and last terms.

The factors of the first term 'fifteen x squared' and the last term 'thirty two' are shown. The product of the first and the last term is four hundred eighty x squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: forty-four x.

After a few trials we see that 5x 5x and 4 are to be multiplied, and 3x 3x and 8 are to be multiplied.

15 x 2 +44x+32=(5x+8)(3x+4) 15 x 2 +44x+32=(5x+8)(3x+4)

Example 4

Factor 18 x 2 −56x+6 18 x 2 −56x+6 .

We see that each term is even, so we can factor out 2.

2(9 x 2 −28x+3) 2(9 x 2 −28x+3)

Notice that the constant term is positive. Thus, we know that the factors of 3 that we are looking for must have the same sign. Since the sign of the middle term is negative, both factors must be negative.
Factor the first and last terms.

 The factors of the first term 'nine x squared' and the last term 'three' are shown. The product of the first and the last term is twenty seven x squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: negative twenty-eight x.

There are not many combinations to try, and we find that 9x 9x and −3 −3 are to be multiplied and x x and −1 −1 are to be multiplied.

18 x 2 −56x+6 = 2(9 x 2 −28x+3) = 2(9x−1)(x−3) 18 x 2 −56x+6 = 2(9 x 2 −28x+3) = 2(9x−1)(x−3)

If we had not factored the 2 out first, we would have gotten the factorization

Eighteen x square minus fifty six x plus six is equal to the product of nine x minus one and two s minus six. Two is common between the terms of the second factor of the polynomial.

The factorization is not complete since one of the factors may be factored further.

18 x 2 −56x+6 = (9x−1)(2x−6) = (9x−1)⋅2(x−3) = 2(9x−1)(x−3) (By the commutative property of multiplication.) 18 x 2 −56x+6 = (9x−1)(2x−6) = (9x−1)⋅2(x−3) = 2(9x−1)(x−3) (By the commutative property of multiplication.)

The results are the same, but it is much easier to factor a polynomial after all common factors have been factored out first.

Example 5

Factor 3 x 2 +x−14 3 x 2 +x−14 .

There are no common factors. We see that the constant term is negative. Thus, the factors of −14 −14 must have different signs.
 Factor the first and last terms.

The factors of the first term 'three x squared' and the last term 'negative fourteen' are shown. The product of the first and the last term is negative forty-two x squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: x.

After a few trials, we see that 3x 3x and −2 −2 are to be multiplied and x x and 7 are to be multiplied.

3 x 2 +x−14=(3x+7)(x−2) 3 x 2 +x−14=(3x+7)(x−2)

Example 6

Factor 8 x 2 −26xy+15 y 2 8 x 2 −26xy+15 y 2 .

We see that the constant term is positive and that the middle term is preceded by a minus sign.
Hence, the factors of 15 y 2 15 y 2 that we are looking for must both be negative.
 Factor the first and last terms.

The factors of the first term 'eight x squared' and the last term 'fifteen y squared' are shown. The product of the first and the last term is one hundred twenty seven x squared y squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: negative twenty-six xy.
After a few trials, we see that 4x 4x and −5y −5y are to be multiplied and 2x 2x and −3y −3y are to be multiplied.

8 x 2 −26xy+15 y 2 =(4x−3y)(2x−5y) 8 x 2 −26xy+15 y 2 =(4x−3y)(2x−5y)

Practice Set A

Factor the following, if possible.

Exercise 1

2 x 2 +13x−7 2 x 2 +13x−7

Solution

(2x−1)(x+7) (2x−1)(x+7)

Exercise 2

3 x 2 +x−4 3 x 2 +x−4

Solution

(3x+4)(x−1) (3x+4)(x−1)

Exercise 3

4 a 2 −25a−21 4 a 2 −25a−21

Solution

(4a+3)(a−7) (4a+3)(a−7)

Exercise 4

16 b 2 −22b−3 16 b 2 −22b−3

Solution

(8b+1)(2b−3) (8b+1)(2b−3)

Exercise 5

10 y 2 −19y−15 10 y 2 −19y−15

Solution

(5y+3)(2y−5) (5y+3)(2y−5)

Exercise 6

6 m 3 +40 m 2 −14m 6 m 3 +40 m 2 −14m

Solution

2m(3m−1)(m+7) 2m(3m−1)(m+7)

Exercise 7

14 p 2 +31pq−10 q 2 14 p 2 +31pq−10 q 2

Solution

(7p−2q)(2p+5q) (7p−2q)(2p+5q)

Exercise 8

−24 w 2 z 2 +14w z 3 −2 z 4 −24 w 2 z 2 +14w z 3 −2 z 4

Solution

−2 z 2 (4w−z)(3w−z) −2 z 2 (4w−z)(3w−z)

Exercise 9

3 x 2 +6xy+2 y 2 3 x 2 +6xy+2 y 2

Solution

not factorable

As you get more practice factoring these types of polynomials you become faster at picking the proper combinations. It takes a lot of practice!
 There is a shortcut that may help in picking the proper combinations. This process does not always work, but it seems to hold true in many cases. After you have factored the first and last terms and are beginning to look for the proper combinations, start with the intermediate factors and not the extreme ones.

Sample Set B

Example 7

Factor 24 x 2 −41x+12 24 x 2 −41x+12 .

Factor the first and last terms.

Table 2
 24 x 2  24 x 2 12
24x,  x 24x,  x −12,  −1 −12,  −1
12x,  2x 12x,  2x   −6,  −2   −6,  −2
  8x,  3x   8x,  3x   −4,  −3   −4,  −3
  6x,  4x   6x,  4x  

Rather than starting with the 24x,x 24x,x and −12,−1 −12,−1 , pick some intermediate values, 8x 8x and 3x 3x , the 6x 6x and 4x 4x , or the −6 −6 and −2 −2 , or the −4 −4 and −3 −3 .

24 x 2 −41x+12=(8x−3)(3x−4) 24 x 2 −41x+12=(8x−3)(3x−4)

Practice Set B

Exercise 10

Factor 48 x 2 +22x-15 48 x 2 +22x-15 .

Solution

(6x+5)(8x−3) (6x+5)(8x−3)

Exercise 11

Factor 54 y 2 +39yw−28 w 2 54 y 2 +39yw−28 w 2 .

Solution

(9y−4w)(6y+7w) (9y−4w)(6y+7w)

The Collect and Discard Method of Factoring ax2+bx+cax2+bx+c

Collect and Discard Method

Consider the polynomial 6 x 2 +x−12 6 x 2 +x−12 . We begin by identifying a a and c c . In this case, a=6 a=6 and c=−12 c=−12 . We start out as we would with a=1 a=1 .

 6 x 2 +x−12:  ( 6x )( 6x ) 6 x 2 +x−12:  ( 6x )( 6x )

Now, compute a⋅c a⋅c .

 a⋅c=(6)(−12)=−72 a⋅c=(6)(−12)=−72

Find the factors of −72 −72 that add to 1, the coefficient of x x , the linear term. The factors are 9 and −8 −8 . Include these factors in the parentheses.

 6 x 2 +x−12:  (6x+9)(6x−8) 6 x 2 +x−12:  (6x+9)(6x−8)

But we have included too much. We must eliminate the surplus. Factor each parentheses.

 6 x 2 +x−12:  3(2x+3)⋅2(3x−4) 6 x 2 +x−12:  3(2x+3)⋅2(3x−4)

Discard the factors that multiply to a=6 a=6 . In this case, 3 and 2. We are left with the proper factorization.

 6 x 2 +x−12=(2x+3)(3x−4) 6 x 2 +x−12=(2x+3)(3x−4)

Sample Set C

Example 8

Factor 10 x 2 +23x−5 10 x 2 +23x−5 .

Identify a=10 a=10 and b=−5 b=−5 .

10 x 2 +23x−5;  ( 10x )( 10x ) 10 x 2 +23x−5;  ( 10x )( 10x )

Compute

a⋅c=(10)(−5)=−50 a⋅c=(10)(−5)=−50

Find the factors of −50 −50 that add to +23 +23 , the coefficient of x x , the linear term. The factors are 25 and −2 −2 . Place these numbers into the parentheses.

10 x 2 +23x−5:  (10x+25)(10x−2) 10 x 2 +23x−5:  (10x+25)(10x−2)

We have collected too much. Factor each set of parentheses and eliminate the surplus.

10 x 2 +23x−5:  (5)(2x+5)⋅(2)(5x−1) 10 x 2 +23x−5:  (5)(2x+5)⋅(2)(5x−1)

Discard the factors that multiply to a=10 a=10 . In this case, 5 and 2.

10 x 2 +23x−5=(2x+5)(5x−1) 10 x 2 +23x−5=(2x+5)(5x−1)

Example 9

Factor 8 x 2 −30x−27 8 x 2 −30x−27 .

Identify a=8 a=8 and c=−27 c=−27 .

8 x 2 −30x−27:  ( 8x )( 8x ) 8 x 2 −30x−27:  ( 8x )( 8x )

Compute

a⋅c=(8)(−27)=−216 a⋅c=(8)(−27)=−216

Find the factors of −216 −216 that add to −30 −30 , the coefficient of x x , the linear term. This requires some thought. The factors are −36 −36 and 6. Place these numbers into the parentheses.

8 x 2 −30x−27:  (8x−36)(8x+6) 8 x 2 −30x−27:  (8x−36)(8x+6)

We have collected too much. Factor each set of parentheses and eliminate the surplus.

8 x 2 −30x−27:  (4)(2x−9)⋅(2)(4x+3) 8 x 2 −30x−27:  (4)(2x−9)⋅(2)(4x+3)

Discard the factors that multiply to a=8 a=8 . In this case, 4 and 2.

8 x 2 −30x−27=(2x−9)(4x+3) 8 x 2 −30x−27=(2x−9)(4x+3)

Example 10

Factor 18 x 2 −5xy−2 y 2 18 x 2 −5xy−2 y 2 .

Identify a=18 a=18 and c=−2 c=−2 .

18 x 2 −5xy−2 y 2 :  ( 18x )( 18x ) 18 x 2 −5xy−2 y 2 :  ( 18x )( 18x )

Compute

a⋅c=(18)(−2)=−36 a⋅c=(18)(−2)=−36

Find the factors of −36 −36 that add to −5 −5 , the coefficient of xy xy . In this case, −9 −9 and 4. Place these numbers into the parentheses, affixing y y to each.

18 x 2 −5xy−2 y 2 :  (18x−9y)(18x+4y) 18 x 2 −5xy−2 y 2 :  (18x−9y)(18x+4y)

We have collected too much. Factor each set of parentheses and eliminate the surplus.

18 x 2 −5xy−2 y 2 :  (9)(2x−y)⋅(2)(9x+2y) 18 x 2 −5xy−2 y 2 :  (9)(2x−y)⋅(2)(9x+2y)

Discard the factors that multiply to a=18 a=18 . In this case, 9 and 4.

18 x 2 −5xy−2 y 2 =(2x−y)(9x+2y) 18 x 2 −5xy−2 y 2 =(2x−y)(9x+2y)

Practice Set C

Exercise 12

Factor 6 x 2 +7x−3 6 x 2 +7x−3 .

Solution

(3x−1)(2x+3) (3x−1)(2x+3)

Exercise 13

Factor 14 x 2 −31x−10 14 x 2 −31x−10 .

Solution

(7x+2)(2x−5) (7x+2)(2x−5)

Exercise 14

Factor 48 x 2 +22x−15 48 x 2 +22x−15 .

Solution

(6x+5)(8x−3) (6x+5)(8x−3)

Exercise 15

Factor 10 x 2 −23xw+12 w 2 10 x 2 −23xw+12 w 2 .

Solution

(5x−4w)(2x−3w) (5x−4w)(2x−3w)

Exercises

Factor the following problems, if possible.

Exercise 16

x 2 +3x+2 x 2 +3x+2

Solution

( x+2 )( x+1 ) ( x+2 )( x+1 )

Exercise 17

x 2 +7x+12 x 2 +7x+12

Exercise 18

2 x 2 +7x+5 2 x 2 +7x+5

Solution

( 2x+5 )( x+1 ) ( 2x+5 )( x+1 )

Exercise 19

3 x 2 +4x+1 3 x 2 +4x+1

Exercise 20

2 x 2 +11x+12 2 x 2 +11x+12

Solution

( 2x+3 )( x+4 ) ( 2x+3 )( x+4 )

Exercise 21

10 x 2 +33x+20 10 x 2 +33x+20

Exercise 22

3 x 2 −x−4 3 x 2 −x−4

Solution

( 3x−4 )( x+1 ) ( 3x−4 )( x+1 )

Exercise 23

3 x 2 +x−4 3 x 2 +x−4

Exercise 24

4 x 2 +8x−21 4 x 2 +8x−21

Solution

( 2x−3 )( 2x+7 ) ( 2x−3 )( 2x+7 )

Exercise 25

2 a 2 −a−3 2 a 2 −a−3

Exercise 26

9 a 2 −7a+2 9 a 2 −7a+2

Solution

not factorable

Exercise 27

16 a 2 +16a+3 16 a 2 +16a+3

Exercise 28

16 y 2 -26y+3 16 y 2 -26y+3

Solution

( 8y−1 )( 2y−3 ) ( 8y−1 )( 2y−3 )

Exercise 29

3 y 2 +14y−5 3 y 2 +14y−5

Exercise 30

10 x 2 +29x+10 10 x 2 +29x+10

Solution

( 5x+2 )( 2x+5 ) ( 5x+2 )( 2x+5 )

Exercise 31

14 y 2 +29y−15 14 y 2 +29y−15

Exercise 32

81 a 2 +19a+2 81 a 2 +19a+2

Solution

not factorable

Exercise 33

24 x 2 +34x+5 24 x 2 +34x+5

Exercise 34

24 x 2 −34x+5 24 x 2 −34x+5

Solution

( 6x−1 )( 4x−5 ) ( 6x−1 )( 4x−5 )

Exercise 35

24 x 2 −26x−5 24 x 2 −26x−5

Exercise 36

24 x 2 +26x−5 24 x 2 +26x−5

Solution

( 6x−1 )( 4x+5 ) ( 6x−1 )( 4x+5 )

Exercise 37

6 a 2 +13a+6 6 a 2 +13a+6

Exercise 38

6 x 2 +5xy+ y 2 6 x 2 +5xy+ y 2

Solution

( 3x+y )( 2x+y ) ( 3x+y )( 2x+y )

Exercise 39

6 a 2 −ay− y 2 6 a 2 −ay− y 2

For the following problems, the given trinomial occurs when solving the corresponding applied problem. Factor each trinomial. You do not need to solve the problem.

Exercise 40

5 r 2 −24r−5 5 r 2 −24r−5 .

It takes 5 hours to paddle a boat 12 miles downstream and then back. The current flows at the rate of 1 mile per hour. At what rate was the boat paddled?

Solution

( 5r+1 )( r−5 ) ( 5r+1 )( r−5 )

Exercise 41

x 2 +5x−84 x 2 +5x−84 .

The length of a rectangle is 5 inches more than the width of the rectangle. If the area of the rectangle is 84 square inches, what are the length and width of the rectangle?

Exercise 42

x 2 +24x−145 x 2 +24x−145 .

A square measures 12 inches on each side. Another square is to be drawn around this square in such a way that the total area is 289 square inches. What is the distance from the edge of the smaller square to the edge of the larger square? (The two squares have the same center.)

Solution

( x+29 )( x−5 ) ( x+29 )( x−5 )

Exercise 43

x 2 +8x−20 x 2 +8x−20 .

A woman wishes to construct a rectangular box that is open at the top. She wishes it to be 4 inches high and have a rectangular base whose length is three times the width. The material used for the base costs $2 per square inch, and the material used for the sides costs $1.50 per square inch. The woman will spend exactly $120 for materials. Find the dimension of the box (length of the base, width of the base, and height).

For the following problems, factor the trinomials if possible.

Exercise 44

16 x 2 −8xy−3 y 2 16 x 2 −8xy−3 y 2

Solution

( 4x+y )( 4x−3y ) ( 4x+y )( 4x−3y )

Exercise 45

6 a 2 +7ab+2 b 2 6 a 2 +7ab+2 b 2

Exercise 46

12 a 2 +7ab+12 b 2 12 a 2 +7ab+12 b 2

Solution

not factorable

Exercise 47

9 x 2 +18xy+8 y 2 9 x 2 +18xy+8 y 2

Exercise 48

8 a 2 +10ab−6 b 2 8 a 2 +10ab−6 b 2

Solution

2( 4 a 2 +5ab−3 b 2 ) 2( 4 a 2 +5ab−3 b 2 )

Exercise 49

12 a 2 +54a−90 12 a 2 +54a−90

Exercise 50

12 b 4 +30 b 2 a+12 a 2 12 b 4 +30 b 2 a+12 a 2

Solution

6( 2 b 2 +a )( b 2 +2a ) 6( 2 b 2 +a )( b 2 +2a )

Exercise 51

30 a 4 b 4 −3 a 2 b 2 −6 c 2 30 a 4 b 4 −3 a 2 b 2 −6 c 2

Exercise 52

3 a 6 −3 a 3 b 2 −18 b 4 3 a 6 −3 a 3 b 2 −18 b 4

Solution

3( a 3 +2 b 2 )( a 3 −3 b 2 ) 3( a 3 +2 b 2 )( a 3 −3 b 2 )

Exercise 53

20 a 2 b 2 +2ab c 2 −6 a 2 c 4 20 a 2 b 2 +2ab c 2 −6 a 2 c 4

Exercise 54

14 a 2 z 2 −40 a 3 z 2 −46 a 4 z 2 14 a 2 z 2 −40 a 3 z 2 −46 a 4 z 2

Solution

2 a 2 z 2 ( 7−20a−23 a 2 ) or −2 a 2 z 2 ( 23 a 2 +20a−7 ) 2 a 2 z 2 ( 7−20a−23 a 2 ) or −2 a 2 z 2 ( 23 a 2 +20a−7 )

Exercises for Review

Exercise 55

((Reference)) Simplify ( a 3 b 6 ) 4 ( a 3 b 6 ) 4 .

Exercise 56

((Reference)) Find the product. x 2 (x−3)(x+4) x 2 (x−3)(x+4) .

Solution

x 4 + x 3 −12 x 2 x 4 + x 3 −12 x 2

Exercise 57

((Reference)) Find the product. (5m−3n) 2 (5m−3n) 2 .

Exercise 58

((Reference)) Solve the equation 5(2x−1)−4(x+7)=0 5(2x−1)−4(x+7)=0 .

Solution

x= 11 2 x= 11 2

Exercise 59

((Reference)) Factor x 5 −8 x 4 +7 x 3 x 5 −8 x 4 +7 x 3 .

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